A space probe has adjustable thrusters. It\'s noticed that if two thrusters (app

Question

A space probe has adjustable thrusters. It's noticed that if two thrusters (applying equal force) are fired for 32.0 s, pushing in the same direction, the probe accelerates from rest and travels some measured distance. How long would it take to move the probe that same distance, if, starting from rest, two thrusters are fired with the same equal forces as before, but are perpendicular to each other? (Probe is in space; all other forces are negligible.) Answer this problem algebraically; do not assign numbers to the problem to work it out that way.

Explanation / Answer

I) F + F = m a1 both in same direction a1 = 2 F / m s = 1/2 a t1^2 = F t1^2 / m distance traveled in first case II) F cos theta + F sin theta = 2 F cos theta = 2^1/2 ^ F = 1.414 F since theta = 45 deg to get the resultant force and sin theta = cos theta = 2^1/2 / 2 = .707 a2 = 2^1/2 F / m = 1.414 F / m s = 1/2 a2 t2^2 = = .707 F t2^2 / m Since the distance is the same in both cases F t1^2 / m = .707 F t2^2 / m t2^2 = (1 / .707) t1^2 Substitute 32 for t1 and solve for t2

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